Solve z^3-z^2+zq=x^2

$$z ^ { 3 } - z ^ { 2 } + z q = x ^ { 2 }$$

Solve for q

$$\left\{\begin{matrix}q=-z^{2}+\frac{x^{2}}{z}+z\text{, }&z\neq 0\\q\in \mathrm{R}\text{, }&x=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract z^{3} from both sides.

$$-z^{2}+zq=x^{2}-z^{3}$$

Add z^{2} to both sides.

$$zq=x^{2}-z^{3}+z^{2}$$

Divide both sides by z.

$$\frac{zq}{z}=\frac{x^{2}-z^{3}+z^{2}}{z}$$

Dividing by z undoes the multiplication by z.

$$q=\frac{x^{2}-z^{3}+z^{2}}{z}$$

Divide x^{2}-z^{3}+z^{2} by z.

$$q=-z^{2}+\frac{x^{2}}{z}+z$$

Solve for x

$$x=\sqrt{z\left(z^{2}-z+q\right)}$$

$$x=-\sqrt{z\left(z^{2}-z+q\right)}\text{, }z\geq \frac{\sqrt{1-4q}+1}{2}\text{ or }q>\frac{1}{4}\text{ or }\left(z=\frac{-\sqrt{1-4q}+1}{2}\text{ and }q<\frac{1}{4}\right)\text{ or }\left(z\geq \frac{-\sqrt{1-4q}+1}{2}\text{ and }q<\frac{1}{4}\text{ and }z\leq 0\right)\text{ or }\left(z\geq 0\text{ and }z\leq \frac{-\sqrt{1-4q}+1}{2}\right)\text{ or }z=0\text{ or }\left(z\geq 0\text{ and }q\geq \frac{1}{4}\right)$$